Q27. - Algebra 2

Question

Use matrices $A$, $B$, $C$and $D$ to find the following.

 

$A=\left[\begin{array}{cc}5& 7\\ -1& 6\\ 3& -9\end{array}\right],B=\left[\begin{array}{cc}8& 3\\ 5& 1\\ 4& 4\end{array}\right],C=\left[\begin{array}{cc}0& 4\\ -2& 5\\ 7& -1\end{array}\right],D=\left[\begin{array}{cc}6& 2\\ 9& 0\\ -3& 0\end{array}\right]$

 

27. $6B-2A$

Step-by-Step Solution

Answer

The final matrix after performing the indicated matrix operations is 

$6B-2A=\left[\begin{array}{cc}38& 4\\ 32& -6\\ 18& \text{42}\end{array}\right]$.

Step 1 - Substitute the matrix

Substitute $\left[\begin{array}{cc}5& 7\\ -1& 6\\ 3& -9\end{array}\right]$ for A and $\left[\begin{array}{cc}8& 3\\ 5& 1\\ 4& 4\end{array}\right]$ for B into $6B-2A$.

 

$6B-2A=6\left[\begin{array}{cc}8& 3\\ 5& 1\\ 4& 4\end{array}\right]-2\left[\begin{array}{cc}5& 7\\ -1& 6\\ 3& -9\end{array}\right]$

Step 2 - Perform the scalar multiplication

Scalar multiplication - The product of a scalar k and an $m\times n$ matrix is an $m\times n$ matrix in which each element equals k times the corresponding elements of the original matrix.

Multiply each element of the first matrix $\left[\begin{array}{cc}8& 3\\ 5& 1\\ 4& 4\end{array}\right]$ by 6 and multiply each element of the second matrix $\left[\begin{array}{cc}5& 7\\ -1& 6\\ 3& -9\end{array}\right]$ by 2.

$\begin{array}{c}6B-2A=6\left[\begin{array}{cc}8& 3\\ 5& 1\\ 4& 4\end{array}\right]-2\left[\begin{array}{cc}5& 7\\ -1& 6\\ 3& -9\end{array}\right]\\ =\left[\begin{array}{cc}6\left(8\right)& 6\left(3\right)\\ 6\left(5\right)& 6\left(1\right)\\ 6\left(4\right)& 6\left(4\right)\end{array}\right]-\left[\begin{array}{cc}2\left(5\right)& 2\left(7\right)\\ 2\left(-1\right)& 2\left(6\right)\\ 2\left(3\right)& 2\left(-9\right)\end{array}\right]\\ =\left[\begin{array}{cc}48& 18\\ 30& 6\\ 24& 24\end{array}\right]-\left[\begin{array}{cc}10& 14\\ -2& 12\\ 6& -18\end{array}\right]\end{array}$

Step 3 - Perform the matrix subtraction

Matrix subtraction - If A and B are two matrices of order $m\times n$, then the subtraction of the two matrices will also yield a matrix of order $m\times n$ wherein each element is the difference of the corresponding elements.

$\begin{array}{c}6B-2A=\left[\begin{array}{cc}48& 18\\ 30& 6\\ 24& 24\end{array}\right]-\left[\begin{array}{cc}10& 14\\ -2& 12\\ 6& -18\end{array}\right]\\ =\left[\begin{array}{cc}48-10& 18-14\\ 30-\left(-2\right)& 6-12\\ 24-6& 24-\left(-18\right)\end{array}\right]\\ =\left[\begin{array}{cc}38& 4\\ 32& -6\\ 18& \text{42}\end{array}\right]\end{array}$